The number of chirps from a Snow Cricket can be used to determine the outside temperature. The number of chirps varies from 10 t
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Answer:
Step-by-step explanation:
Since we're finding the product, we have to multiply:
×
You can simplify in this stage by using the "butterfly method", and dividing by
, and
by
, you'd then have:
×
Multiply the numerators and the denominators to get:
~
If you prefer the longer way, again, multiply:
×
Multiply the numerators and the denominators:
Simplify the fraction by dividing both the numerator and denominator by :
Answer:
(a) Null Hypothesis, :
or
Alternate Hypothesis, :
or
(b) We conclude that there is a statistical difference in these two proportions measured initially and then one year later.
Step-by-step explanation:
We are given that a random sample of 1,250 drivers found that 98.9% were in compliance. A year after the implementation, compliance was again measured to see if compliance was the same (or not) as previously measured.
A different random sample of 1,100 drivers found 96.9% compliance."
<em />
<em>Let </em><em> = proportion of drivers that were in compliance initially</em>
<em> = proportion of drivers that were in compliance one year later</em>
(a) <u>Null Hypothesis</u>, :
or
{means that there is not any statistical difference in these two proportions measured initially and then one year later}
<u>Alternate Hypothesis</u>, :
or
{means that there is a statistical difference in these two proportions measured initially and then one year later}
The test statistics that will be used here is <u>Two-sample z proportion statistics</u>;
T.S. = ~ N(0,1)
where, = sample proportion of drivers in compliance initially = 98.9%
= sample proportion of drivers in compliance one year later = 96.9%
= sample of drivers initially = 1,250
= sample of drivers one year later = 1,100
(b) So, <u><em>the test statistics</em></u> =
= 3.33
<u>Now, P-value of the test statistics is given by;</u>
P-value = P(Z > 3.33) = 1 - P(Z 3.33)
= 1 - 0.99957 = <u>0.00043</u>
Since in the question we are not given with the level of significance so we assume it to be 5%. Now at 5% significance level, the z table gives critical values between -1.96 and 1.96 for two-tailed test.
<em>Since our test statistics does not lies within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which </em><u><em>we reject our null hypothesis.</em></u>
Therefore, we conclude that there is a statistical difference in these two proportions measured initially and then one year later.